= Airy disk derivation
{c}
{title2=$A(u)=2J_1(u)/u$}
Write $a=D/2$ and <wavenumber> $k=2\pi/\lambda$. The pupil <Fourier transform>, normalized by pupil <area>, gives the <Fraunhofer diffraction> <wave amplitude>. In pupil <polar coordinates>,
$$
A(u)=\frac1{\pi a^2}\int_0^a\int_0^{2\pi}e^{-ik\rho\sin\theta\cos\phi}\rho\,d\phi\,d\rho=2\int_0^1tJ_0(ut)\,dt,\qquad u=ka\sin\theta.
$$
The angular average is $J_0$ because $J_0(z)=(2\pi)^{-1}\int_0^{2\pi}e^{-iz\cos\phi}\,d\phi$. The <Bessel function of the first kind> identity $d\{zJ_1(z)\}/dz=zJ_0(z)$ then reduces the radial <integral> to $A(u)=2J_1(u)/u$. Taking the squared magnitude yields the <Airy disk> <light intensity> profile.<Oscillations and Waves by Richard Fitzpatrick/Two-Dimensional Fourier Optics>[equations 10.78–10.84: angular integral, radial integral and Bessel identities]{ref}
The apparent division by zero is removable. The <power series>
$$
J_1(u)=\frac u2-\frac{u^3}{16}+\frac{u^5}{384}+O(u^7)
$$
gives $A(u)=1-u^2/8+u^4/192+O(u^6)$ and $P(u)=1-u^2/4+5u^4/192+O(u^6)$. At the centre, every pupil contribution has the same <wave phase>, so their normalized sum is one. The first dark ring instead corresponds to $J_1(u)=0$ with $u>0$, rather than to the removable zero of $J_1$ at the origin.
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